Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = int(1e5) + 23;
int up[N][20];
int d[N][20][12], sz[N][20], dp[N][20][12];
int n, u, v;
int tp[N][12], ss[N];
vector<int> g[N];
int c[N];
vector<int> rb[N];
int q, a;
bool used[N];
int tmp[20], tn;
int t_in[N], t_out[N];
int t;
int ans[N], an;
int bn, b[N], cn;
int m;
void upd1(int v, int c) {
  int k = ss[v];
  if (k == 0) {
    tp[v][++k] = c;
    ss[v] = k;
    return;
  }
  tn = 0;
  for (int i = 1; i <= k; i++) {
    if (c < tp[v][i]) {
      tmp[++tn] = c;
      for (int j = i; j <= k; j++) {
        tmp[++tn] = tp[v][j];
      }
      break;
    }
    tmp[++tn] = tp[v][i];
  }
  if (tn <= k) tmp[++tn] = c;
  tn = min(tn, 10);
  ss[v] = tn;
  for (int i = 1; i <= tn; i++) {
    tp[v][i] = tmp[i];
  }
}
void upd(int v, int u, int x, int y) {
  int k1 = ss[v];
  int k2 = ss[u];
  int j = 1;
  tn = 0;
  int i = 1;
  while (i <= k1) {
    if (j > k2 || tp[v][i] < tp[u][j]) {
      tmp[++tn] = tp[v][i];
      i++;
    } else if (tp[u][j] < tp[v][i]) {
      tmp[++tn] = tp[u][j];
      j++;
    } else {
      if (tp[u][j] == tp[v][i]) {
        j++;
        i++;
        tmp[++tn] = tp[u][j];
      }
    }
  }
  while (j <= k2) {
    tmp[++tn] = tp[u][j];
    j++;
  }
  tn = min(tn, 10);
  sz[x][y] = tn;
  for (int i = 1; i <= tn; i++) {
    dp[x][y][i] = tmp[i];
  }
}
void upd2(int v, int u, int r, int x, int y) {
  int k1 = sz[v][r];
  int k2 = sz[u][r];
  int j = 1, i = 1;
  tn = 0;
  while (i <= k1) {
    if (j > k2 || dp[v][r][i] < dp[u][r][j]) {
      tmp[++tn] = dp[v][r][i];
      i++;
    } else if (dp[u][r][j] < dp[v][r][i]) {
      tmp[++tn] = dp[u][r][j];
      j++;
    } else if (dp[u][r][j] == dp[v][r][i]) {
      tmp[++tn] = dp[u][r][j];
      j++;
      i++;
    }
  }
  while (j <= k2) {
    tmp[++tn] = dp[u][r][j];
    j++;
  }
  tn = min(tn, 10);
  sz[x][y] = tn;
  for (int i = 1; i <= tn; i++) {
    dp[x][y][i] = tmp[i];
  }
}
void dfs(int v) {
  used[v] = 1;
  t_in[v] = ++t;
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (!used[to]) {
      up[to][0] = v;
      dfs(to);
    }
  }
  t_out[v] = ++t;
}
bool ok(int u, int v) { return (t_in[u] <= t_in[v] && t_out[u] >= t_out[v]); }
void build_lca() {
  for (int i = 1; i <= n; i++) {
    if (up[i][0] != -1) upd(i, up[i][0], i, 0);
  }
  for (int j = 1; j <= 17; j++) {
    for (int i = 1; i <= n; i++)
      if (up[i][j - 1] != -1) {
        up[i][j] = up[up[i][j - 1]][j - 1];
        upd2(i, up[i][j - 1], j - 1, i, j);
      }
  }
}
int lca(int v, int u) {
  if (ok(v, u)) return v;
  if (ok(u, v)) return u;
  for (int i = 17; i >= 0; i--) {
    if (up[v][i] != -1 && !ok(up[v][i], u)) {
      v = up[v][i];
    }
  }
  return up[v][0];
}
void Merge(int t, int v, int u) {
  if (t == 1) {
    int k = sz[v][u];
    int j = 1, i = 1;
    tn = 0;
    while (i <= k) {
      if (j > cn || dp[v][u][i] < c[j]) {
        tmp[++tn] = dp[v][u][i];
        i++;
      } else if (dp[v][u][i] > c[j]) {
        tmp[++tn] = c[j];
        j++;
      } else if (dp[v][u][i] == c[j]) {
        tmp[++tn] = c[j];
        i++;
        j++;
      }
    }
    while (j <= cn) {
      tmp[++tn] = c[j];
      j++;
    }
    tn = min(tn, 10);
    for (int i = 1; i <= tn; i++) {
      c[i] = tmp[i];
    }
    cn = tn;
  } else {
    int k = sz[v][u];
    int j = 1, i = 1;
    tn = 0;
    while (i <= k) {
      if (j > bn || dp[v][u][i] < b[j]) {
        tmp[++tn] = dp[v][u][i];
        i++;
      } else if (dp[v][u][i] > b[j]) {
        tmp[++tn] = b[j];
        j++;
      } else if (dp[v][u][i] == b[j]) {
        tmp[++tn] = b[j];
        j++;
        i++;
      }
    }
    while (j <= bn) {
      tmp[++tn] = b[j];
      j++;
    }
    tn = min(tn, 10);
    for (int i = 1; i <= tn; i++) {
      b[i] = tmp[i];
    }
    bn = tn;
  }
}
void way(int v, int u, int t) {
  if (t == 1) {
    int k = ss[v];
    if (u == v) {
      for (int i = 1; i <= k; i++) {
        c[++cn] = tp[v][i];
      }
      return;
    }
    for (int i = 17; i >= 0; i--) {
      if (up[v][i] != -1 && !ok(up[v][i], u)) {
        Merge(1, v, i);
        v = up[v][i];
      }
    }
    Merge(1, v, 0);
  } else {
    int k = ss[v];
    if (u == v) {
      for (int i = 1; i <= k; i++) {
        b[++bn] = tp[v][i];
      }
      return;
    }
    for (int i = 17; i >= 0; i--) {
      if (up[v][i] != -1 && !ok(up[v][i], u)) {
        Merge(0, v, i);
        v = up[v][i];
      }
    }
    Merge(0, v, 0);
  }
}
void solve() {
  cn = 0;
  bn = 0;
  int x = lca(u, v);
  way(u, x, 1);
  way(v, x, 0);
  tn = 0;
  int j = 1, i = 1;
  while (i <= cn) {
    if (j > bn || c[i] < b[j]) {
      tmp[++tn] = c[i];
      i++;
    } else if (b[j] < c[i]) {
      tmp[++tn] = b[j];
      j++;
    } else if (b[j] == c[i]) {
      tmp[++tn] = b[j];
      j++;
      i++;
    }
  }
  while (j <= bn) {
    tmp[++tn] = b[j];
    j++;
  }
  tn = min(tn, a);
  printf("%d ", tn);
  for (int i = 1; i <= tn; i++) {
    printf("%d ", tmp[i]);
  }
  puts("");
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &u, &v);
    g[v].push_back(u);
    g[u].push_back(v);
  }
  memset(up, -1, sizeof(up));
  dfs(1);
  for (int i = 1; i <= m; i++) {
    scanf("%d", &c[i]);
    upd1(c[i], i);
  }
  build_lca();
  for (int i = 1; i <= q; i++) {
    scanf("%d%d%d", &u, &v, &a);
    solve();
  }
  return 0;
}